Question

A certain system can experience three different types of defects. Let \({{\rm{A}}_{\rm{1}}}\)\({\rm{(i = 1,2,3)}}\)denote the event that the system has a defect of type Suppose that

\(\begin{aligned}P\left( {{A_1}} \right) &= .12\;\;\;P\left( {{A_2}} \right) = .07\;\;\;P\left( {{A_3}}\right) = .05 \hfill \\P\left( {{A_1}E {A_2}} \right) = .13\;\;\;P\left( {{A_1}E {A_3}}\right) &= .14 \hfill \\P\left( {{A_2}E {A_2}} \right) = .10\;\;\;P\left( {{A_1}{C}{A_3}{C}{A_7}} \right) &= .01 \hfill \\\end{aligned} \)

a. What is the probability that the system does not have a type 1 defect?

b. What is the probability that the system has both type 1 and type 2 defects?

c. What is the probability that the system has both type 1 and type 2 defects but not a type 3 defect?

d. What is the probability that the system has at most two of these defects?

Short answer

1. \({\rm{P}}\left( {{\rm{not}}{{\rm{A}}_{\rm{1}}}} \right){\rm{ = 0}}{\rm{.88 = 88\% }}\)
2. \({\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{{C}}}{{\rm{A}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.06 = 6\% }}\)
3. \({\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{{C}}}{{\rm{A}}_{\rm{2}}}{\rm{{C}A}}_{\rm{3}}^{\rm{c}}} \right){\rm{ =0}}{\rm{.05 = 5\% }}\)
4. \({\rm{P(}}\)At most 2 defect) \({\rm{) = 0}}{\rm{.99 = 99\% }}\)

Step-by-step solution

Definition of Probability and statistics

Probability refers to the likelihood of a random event's outcome. This word refers to determining the likelihood of a given occurrence occurring. What are the chances of receiving a head when we toss a coin in the air, for example? The amount of alternative outcomes determines the answer to this question. In this case, the conclusion might be either head or tail. As a result, the chance of getting a head is 1/2.

The study of data collection, analysis, interpretation, presentation, and organization is known as statistics. It is a process of gathering and analyzing data. This has a wide range of uses, from little to huge. Stats are employed in every data analysis, whether it is the study of the country's population or its economy.

calculating the probability that the system does not have a type 1 defect

Given:

\(\begin{aligned}{{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ = 0}}{\rm{.12}}}\\{{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.07}}}\\{{\rm{P}}\left( {{{\rm{A}}_{\rm{3}}}} \right){\rm{ = 0}}{\rm{.05}}}\\{{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{E }}{{\rm{A}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.13}}}\\{{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{E }}{{\rm{A}}_{\rm{3}}}} \right){\rm{ = 0}}{\rm{.14}}}\\{{\rm{P}}\left( {{{\rm{A}}_{\rm{2}}}{\rm{E }}{{\rm{A}}_{\rm{3}}}} \right){\rm{ = 0}}{\rm{.10}}}\\{{\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{{C}}}{{\rm{A}}_{\rm{2}}}{\rm{{C}}}{{\rm{A}}_{\rm{3}}}} \right){\rm{ = 0}}{\rm{.01}}}\end{aligned}\)

The complement rule is as follows.The occurrence \({{\rm{A}}_{\rm{1}}}\) refers to a kind 1 defect.

\({\rm{P}}\left( {{\rm{not}}{{\rm{A}}_{\rm{1}}}} \right){\rm{ = 1 - P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ = 1 - 0}}{\rm{.12 = 0}}{\rm{.88 = 88\% }}\)

calculating the probability that the system has both type 1 and type 2 defects

The following is the general addition rule for any two events:

\({\rm{P(A{C}B) = P(A) + P(B) - P(A E B)}}\)

A type 1 flaw is classified as \({{\rm{A}}_{\rm{1}}}\), whereas a type 2 defect is classified as \({{\rm{A}}_{\rm{2}}}\)\({\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{{C}}}{{\rm{A}}_{\rm{2}}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{\rm{1}}}} \right){\rm{ + P}}\left( {{{\rm{A}}_{\rm{2}}}} \right){\rm{ - P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{E }}{{\rm{A}}_{\rm{2}}}} \right){\rm{ = 0}}{\rm{.12 + 0}}{\rm{.07 - 0}}{\rm{.13 = 0}}{\rm{.06 = 6\% }}\)

Calculating the probability that the system has both type 1 and type 2 defects but not a type 3 defect

If the system is in the intersection of \({{\rm{A}}_{\rm{1}}}\) and \({{\rm{A}}_{\rm{2}}}\)but not in the intersection of all three events \({{\rm{A}}_{\rm{i}}}\), the system has both type 1 and type 2 faults but no type 3 defect.

\({\rm{P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{{C}}}{{\rm{A}}_{\rm{2}}}{\rm{{C}A}}_{\rm{3}}^{\rm{c}}} \right){\rm{ = P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{{C}}}{{\rm{A}}_{\rm{2}}}} \right){\rm{ - P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{{C}}}{{\rm{A}}_{\rm{2}}}{\rm{{C}}}{{\rm{A}}_{\rm{3}}}} \right){\rm{ = 0}}{\rm{.06 - 0}}{\rm{.01 = 0}}{\rm{.05 = 5\% }}\)

Calculating the probability that the system has has at most two of these defects

The complement rule is as follows:

\({\rm{P(notA) = 1 - P(A)}}\)

If the system does not fall in the intersection of all three occurrences, it has at least two flaws.\({{\rm{A}}_{\rm{i}}}\)\({\rm{P(\;At most\;2\;defects\;) = 1 - P}}\left( {{{\rm{A}}_{\rm{1}}}{\rm{{C}}}{{\rm{A}}_{\rm{2}}}{\rm{{C}}}{{\rm{A}}_{\rm{3}}}} \right){\rm{ = 1 - 0}}{\rm{.01 = 0}}{\rm{.99 = 99\% }}\)